Higher Outcome 2 Higher Unit 1 What is a set Recognising a Function in various formats Composite Functions Exponential and Log Graphs Connection between Radians and degrees & Exact values Solving Trig Equations Basic Trig Identities Exam Type Questions Graph Transformations Trig Graphs

Higher Outcome 2 Sets & Functions Notation & Terminology SETS: A set is a collection of items which have some common property. These items are called the members or elements of the set. Sets can be described or listed using “curly bracket” notation.

Higher Outcome 2 eg{colours in traffic lights} eg {square nos. less than 30} DESCRIPTION LIST NB: Each of the above sets is finite because we can list every member = {red, amber, green} = { 0, 1, 4, 9, 16, 25} Sets & Functions

Higher Outcome 2 Sets & Functions N = {natural numbers} = {1, 2, 3, 4, ……….} W = {whole numbers} = {0, 1, 2, 3, ………..} Z = {integers}= {….-2, -1, 0, 1, 2, …..} Q = {rational numbers} This is the set of all numbers which can be written as fractions or ratios. eg 5 = 5 / 1 -7 = -7 / = 6 / 10 = 3 / 5 55% = 55 / 100 = 11 / 20 etc We can describe numbers by the following sets:

Higher Outcome 2 R = {real numbers} This is all possible numbers. If we plotted values on a number line then each of the previous sets would leave gaps but the set of real numbers would give us a solid line. We should also note that N “fits inside” W W “fits inside” Z Z “fits inside” Q Q “fits inside” R Sets & Functions

Higher Outcome 2 Sets & Functions QZWN When one set can fit inside another we say that it is a subset of the other. The members of R which are not inside Q are called irrational numbers. These cannot be expressed as fractions and include ,  2, 3  5 etc R

Higher Outcome 2 To show that a particular element/number belongs to a particular set we use the symbol . eg 3  W but 0.9  Z Examples { x  W: x < 5 }= { 0, 1, 2, 3, 4 } { x  Z: x  -6 }= { -6, -5, -4, -3, -2, …….. } { x  R: x 2 = -4 }= { } or  This set has no elements and is called the empty set. Sets & Functions

Higher Outcome 2 If the first set is A and the second B then we often write f: A  B Functions & Mappings Defn: A function or mapping is a relationship between two sets in which each member of the first set is connected to exactly one member in the second set. The members of set A are usually referred to as the domain of the function (basically the starting values or even x-values) while the corresponding values or images come from set B and are called the range of the function (these are like y-values).

Higher Outcome 2 Functions & Mapping Functions can be illustrated in three ways: 1) by a formula. 2) by arrow diagram. 3) by a graph (ie co-ordinate diagram). Example Suppose that f: A  B is defined by f(x) = x 2 + 3x where A = { -3, -2, -1, 0, 1}. FORMULA then f(-3) = 0,f(-2) = -2, f(-1) = -2, f(0) = 0, f(1) = 4 NB: B = {-2, 0, 4} = the range!

Higher Outcome A B ARROW DIAGRAM Functions & Mapping f(-3) = 0 f(-2) = -2 f(-1) = -2 f(0) = 0 f(1) = 4 f(x)

Higher Outcome 2 Functions & Graphs In a GRAPH we get : NB: This graph consists of 5 separate points. It is not a solid curve.

Higher Outcome 2 Recognising Functions ABAB abcdabcd efgefg Not a function two arrows leaving b! ABAB abcdabcd e f ge f g YES Functions & Graphs

Higher Outcome 2 Functions & Graphs A B abcdabcd efgefg Not a function - d unused! ABAB abcdabcd efghefgh YES

Higher Outcome 2 Functions & Graphs Recognising Functions from Graphs If we have a function f: R  R (R - real nos.) then every vertical line we could draw would cut the graph exactly once! This basically means that every x-value has one, and only one, corresponding y-value!

Higher Outcome 2 Function & Graphs x Y Function !!

Higher Outcome 2 x Y Not a function !! Cuts graph more than once ! Function & Graphs x must map to one value of y

Higher Outcome 2 Functions & Graphs X Y Not a function !! Cuts graph more than once!

Higher Outcome 2 X Y Function !! Functions & Graphs

Higher Outcome 1 COMPOSITION OF FUNCTIONS ( or functions of functions ) Suppose that f and g are functions where f:A  B and g:B  C with f(x) = y and g(y) = z where x  A, y  B and z  C. Suppose that h is a third function where h:A  C with h(x) = z. Composite Functions

Higher Outcome 1 Composite Functions ie ABCABC x y z f g h We can say that h(x) = g(f(x)) “function of a function”

Higher Outcome 1 Composite Functions f(2)=3 x 2 – 2 =4 g(4)= =17 f(5)=5x3-2 =13 Example 1 Suppose that f(x) = 3x - 2 and g(x) = x 2 +1 (a) g( f(2) ) =g(4) = 17 (b) f( g (2) ) = f(5) = 13 (c) f( f(1) ) =f(1)= 1 (d) g( g(5) )= g(26)= 677 f(1)=3x1 - 2 =1 g(26)= =677 g(2)= =5 f(1)=3x1 - 2 =1 g(5)= =26

Higher Outcome 1 Suppose that f(x) = 3x - 2 and g(x) = x 2 +1 Find formulae for (a) g(f(x)) (b) f(g(x)). (a) g(f(x)) = g(3x-2)= (3x-2) 2 + 1= 9x x + 5 (b) f(g(x)) = f(x 2 + 1)= 3(x 2 + 1) - 2= 3x CHECK g(f(2)) =9 x x 2 + 5= = 17 f(g(2)) =3 x = 13As in Ex1 NB: g(f(x))  f(g(x)) in general. Composite Functions

Higher Outcome 1 Let h(x) = x - 3, g(x) = x and k(x) = g(h(x)). If k(x) = 8 then find the value(s) of x. k(x) = g(h(x)) = g(x - 3) = (x - 3) = x 2 - 6x + 13 Put x 2 - 6x + 13 = 8 then x 2 - 6x + 5 = 0 or (x - 5)(x - 1) = 0 So x = 1 or x = 5 CHECKg(h(5)) = g(2)= = 8 Composite Functions

Higher Outcome 1 Choosing a Suitable Domain (i) Suppose f(x) = 1. x Clearly x  0 So x 2  4 So x  -2 or 2 Hence domain = {x  R: x  -2 or 2 } Composite Functions

Higher Outcome 1 (ii) Suppose that g(x) =  (x 2 + 2x - 8) We need (x 2 + 2x - 8)  0 Suppose (x 2 + 2x - 8) = 0 Then (x + 4)(x - 2) = 0 So x = -4 or x = 2 So domain = { x  R: x  -4 or x  2 } Composite Functions Check values below -4, between -4 and 2, then above 2 x = -5 (-5 + 4)(-5 - 2) = positive x = 0 (0 + 4)(0 - 2) = negative x = 3 (3 + 4)(3 - 2) = positive -42

Higher Outcome 1 A function in the form f(x) = a x where a > 0, a ≠ 1 is called an exponential function to base a. Exponential (to the power of) Graphs Exponential Functions Consider f(x) = 2 x x f(x) 11 / 8 ¼ ½

Higher Outcome 1 The graph is like y = 2 x (0,1) (1,2) Major Points (i) y = 2 x passes through the points (0,1) & (1,2) (ii) As x  ∞ y  ∞ however as x  ∞ y  0. (iii) The graph shows a GROWTH function. Graph

Higher Outcome 1 ie y x 1 / 8 ¼ ½ To obtain y from x we must ask the question “What power of 2 gives us…?” This is not practical to write in a formula so we say y = log 2 x “the logarithm to base 2 of x” or “log base 2 of x” Log Graphs

Higher Outcome 1 The graph is like y = log 2 x (1,0) (2,1) Major Points (i) y = log 2 x passes through the points (1,0) & (2,1). (ii) As x  y  but at a very slow rate and as x  0 y  - . NB: x > 0 Graph

Higher Outcome 1 The graph of y = a x always passes through (0,1) & (1,a) It looks like.. x Y y = a x (0,1) (1,a) Exponential (to the power of) Graphs

Higher Outcome 1 The graph of y = log a x always passes through (1,0) & (a,1) It looks like.. x Y y = log a x (1,0) (a,1) Log Graphs

Higher Outcome 1 Graph Transformations We will use the TI – 83 to investigate f(x) graphs of the form 1.f(x) ± k 2.f(x ± k) 3.-f(x) 4.f(-x) 5.kf(x) 6.f(kx) Each moves the Graph of f(x) in a certain way !

Higher Outcome 1 y = f(x)y = x 2 y = x 2 -3 Mathematically y = f(x) ± k moves f(x) up or down Depending on the value of k + k  move up - k  move down y = x y = f(x) ± k Graph of f(x) ± k Transformations

Higher Outcome 1 Graph of -f(x ± k) Transformations y = f(x)y = x 2 y = (x-1) 2 y = (x+2) 2 y = f(x ± k) Mathematically y = f(x ± k) moves f(x) to the left or right depending on the value of k -k  move right + k  move left

Higher Outcome 1 y = f(x)y = x 2 y = -x 2 Mathematically y = –f(x) reflected f(x) in the x - axis y = -f(x) Graph of -f(x) Transformations

Higher Outcome 1 y = f(x)y = 2x + 3 y = -(2x + 3) Mathematically y = –f(x) reflected f(x) in the x - axis y = -f(x) Graph of -f(x) Transformations

Higher Outcome 1 y = f(x)y = x 3 y = -x 3 Mathematically y = –f(x) reflected f(x) in the x - axis y = -f(x) Graph of -f(x) Transformations

Higher Outcome 1 Graph of f(-x) Transformations y = f(x)y = x + 2 y = -x + 2 Mathematically y = f(-x) reflected f(x) in the y - axis y = f(-x)

Higher Outcome 1 y = f(x)y = (x+2) 2 y = (-x+2) 2 Mathematically y = f(-x) reflected f(x) in the y - axis y = f(-x) Graph of f(-x) Transformations

Higher Outcome 1 Graph of k f(x) Transformations y = f(x)y = x 2 -1 y = 4(x 2 -1) Mathematically y = k f(x) Multiply y coordinate by a factor of k k > 1  (stretch in y-axis direction) 0 < k < 1  ( squash in y-axis direction) y = 0.25(x 2 -1) y = k f(x)

Higher Outcome 1 y = f(x)y = x 2 -1 y = -4(x 2 -1) Mathematically y = -k f(x) k = -1 reflect graph in x-axis k < -1  reflect f(x) in x-axis & multiply by a factor k (stretch in y-axis direction) 0 < k < -1  reflect f(x) in x-axis multiply by a factor k (squash in y-axis direction) y = -0.25(x 2 -1) y = k f(x) Graph of -k f(x) Transformations

Higher Outcome 1 y = f(x)y = (x-2) 2 y = (2x-2) 2 Mathematically y = f(kx) Multiply x – coordinates by 1/k k > 1  squashes by a factor of 1/k in the x-axis direction k < 1  stretches by a factor of 1/k in the x-axis direction y = (0.5x-2) 2 y = f(kx) Graph of f(kx) Transformations

Higher Outcome 1 y = f(x)y = (x-2) 2 y = (-2x-2) 2 Mathematically y = f(-kx) k = -1 reflect in y-axis k < -1  reflect & squashes by factor of 1/k in x direction -1 < k > 0  reflect & stretches factor of 1/k in x direction y = (-0.5x-2) 2 y = f(-kx) Graph of f(-kx) Transformations

Higher Outcome 1 Trig Graphs The same transformation rules apply to the basic trig graphs. NB: If f(x) =sinx  then 3f(x) = 3sinx  and f(5x) = sin5x  Think about sin replacing f ! Also if g(x) = cosx  then g(x) –4 = cosx  –4 and g(x+90) = cos(x+90)  Think about cos replacing g !

Higher Outcome 1 Sketch the graph of y = sinx  - 2 If sinx  = f(x) then sinx  - 2 = f(x) - 2 So move the sinx  graph 2 units down. y = sinx  - 2 Trig Graphs

Higher Outcome 1 Sketch the graph of y = cos(x - 50)  If cosx  = f(x) then cos(x - 50)  = f(x- 50) So move the cosx  graph 50 units right. y = cos(x - 50)  Trig Graphs

Higher Outcome 1 Trig Graphs Sketch the graph of y = 3sinx  If sinx  = f(x) then 3sinx  = 3f(x) So stretch the sinx  graph 3 times vertically. y = 3sinx 

Higher Outcome 1 Trig Graphs Sketch the graph of y = cos4x  If cosx  = f(x) then cos4x  = f(4x) So squash the cosx  graph to 1 / 4 size horizontally y = cos4x 

Higher Outcome 1 Trig Graphs Sketch the graph of y = 2sin3x  If sinx  = f(x) then 2sin3x  = 2f(3x) So squash the sinx  graph to 1 / 3 size horizontally and also double its height. y = 2sin3x 

Higher Outcome 1 Radians Radian measure is an alternative to degrees and is based upon the ratio of arc length radius r θ a θ- theta θ(radians) =

Higher Outcome 1 If the arc length = the radius θ r a = r θ (radians) = r / r = 1 If we now take a semi-circle θ (180 o ) r a Here a = ½ of circumference = ½ of πd = πr So θ (radians) = πr / r = π Radians

Higher Outcome 1 Converting degrees radians ÷180 then X π ÷ π then x 180

Higher Outcome 1 Since we have a semi-circle the angle must be 180 o. We now get a simple connection between degrees and radians. π (radians) = 180 o This now gives us 2π = 360 o π / 2 = 90 o π / 3 = 60 o 3π / 2 = 270 o 2π / 3 = 120 o π / 4 = 45 o 3π / 4 = 135 o π / 6 = 30 o 5π / 6 = 150 o NB: Radians are usually expressed as fractional multiples of π. Radians

Higher Outcome 1 Ex172 o = 72 / 180 X π =2π / 5 Ex2330 o = 330 / 180 X π = 11 π / 6 Ex32π / 9 =2π / 9 ÷ π x 180 o = 2 / 9 X 180 o =40 o Ex4 23π/ 18 =23π / 18 ÷ π x 180 o = 23 / 18 X 180 o =230 o Converting

Higher Outcome º º 33 This triangle will provide exact values for sin, cos and tan 30º and 60º Exact Values Some special values of Sin, Cos and Tan are useful left as fractions, We call these exact values

Higher Outcome 1 x0º30º45º60º90º Sin xº Cos xº Tan xº  ½ ½ 33  3 2  Exact Values

Higher Outcome 1 Exact Values 11 45º 2 2 Consider the square with sides 1 unit 1 1 We are now in a position to calculate exact values for sin, cos and tan of 45 o

Higher Outcome 1 x0º30º45º60º90º Sin xº Cos xº Tan xº  ½ ½ 33  3 2  Exact Values 1  2 1

Higher Outcome 1 Exact value table and quadrant rules. tan150 o (Q2 so neg) = tan(180-30) o = -tan30 o = -1 / √3 cos300 o (Q4 so pos) = cos(360-60) o = cos60 o = 1 / 2 sin120 o (Q2 so pos) = sin(180-60) o = sin60 o = √ 3/2 tan300 o (Q4 so neg) = tan(360-60) o = -tan60 o = - √ 3

Higher Outcome 1 Find the exact value of cos 2 ( 5π / 6 ) – sin 2 ( π / 6 ) cos( 5π / 6 ) = cos150 o (Q2 so neg) = cos(180-30) o = -cos30 o = - √3 / 2 sin( π / 6 )= sin30 o = 1 / 2 cos 2 ( 5π / 6 ) – sin 2 ( π / 6 )= (- √3 / 2 ) 2 – ( 1 / 2 ) 2 = ¾ - 1 / 4 = 1 / 2 Exact value table and quadrant rules.

Higher Outcome 1 Exact value table and quadrant rules. Prove thatsin( 2 π / 3 ) = tan ( 2 π / 3 ) cos ( 2 π / 3 ) sin( 2π / 3 ) = sin120 o = sin(180 – 60) o = sin60 o = √3 / 2 cos( 2 π / 3 ) = cos120 o tan( 2 π / 3 ) = tan120 o = cos(180 – 60) o = tan(180 – 60) o = -cos60 o = -tan60 o = - 1 / 2 = - √3 LHS = sin( 2 π / 3 ) cos ( 2 π / 3 ) = √ 3 / 2 ÷ - 1 / 2 = √3 / 2 X -2 = - √3= tan( 2π / 3 )= RHS

Higher Outcome 2 Solving Trig Equations All +veSin +ve Tan +ve Cos +ve 180 o - x o 180 o + x o 360 o - x o 1234

Higher Outcome 2 Solving Trig Equations Example 1 Type 1: Solving the equation sin x o = 0.5 in the range 0 o to 360 o Graphically what are we trying to solve x o = sin -1 (0.5) x o = 30 o There is another solution x o = 150 o (180 o – 30 o = 150 o ) sin x o = (0.5) 1234 C A S T 0o0o 180 o 270 o 90 o

Higher Outcome 2 Solving Trig Equations Example 2 : Solving the equation cos x o = 0 in the range 0 o to 360 o Graphically what are we trying to solve cos x o = x o = 51.3 o (360 o o = o ) x o = cos -1 (0.625) There is another solution 1234 x o = o C A S T 0o0o 180 o 270 o 90 o

Higher Outcome 2 Solving Trig Equations Example 3 : Solving the equation tan x o – 2 = 0 in the range 0 o to 360 o Graphically what are we trying to solve tan x o = 2 x o = 63.4 o x = 180 o o = o x o = tan -1 (2) There is another solution 1234 C A S T 0o0o 180 o 270 o 90 o

Higher Outcome 2 Solving Trig Equations Example 4 Type 2 : Solving the equation sin 2x o = 0 in the range 0 o to 360 o Graphically what are we trying to solve 2x o = sin -1 (0.6) 2x o =217 o, 323 o 577 o, 683 o sin 2x o = (-0.6) x o =108.5 o, o o, o C A S T 0o0o 180 o 270 o 90 o 2x o = 37 o ( always 1 st Q First) ÷2

Higher Outcome 2 created by Mr. Lafferty Solving Trig Equations Graphically what are we trying to solve sin (2x +30 o ) = √3 ÷ 2 2x o - 30 o = 60 o, 120 o,420 o, 480 o sin (2x + 30 o ) = √3 x o =45 o, 75 o 225 o, 265 o 2x - 30 o = sin -1 (√3 ÷ 2) Example 5 Type 3 : Solving the equation 2sin (2x o - 30 o ) - √3 = 0 in the range 0 o to 360 o C A S T 0o0o 180 o 270 o 90 o 2x o = 90 o, 150 o,450 o, 530 o ÷2

Higher Outcome 2 created by Mr. Lafferty Solving Trig Equations Example 5 Type 4 : Solving the equation cos 2 x = 1 in the range 0 o to 360 o Graphically what are we trying to solve cos x o = ± 1 cos x o = 1 cos 2 x o = 1 x o = 0 o and 360 o C A S T 0o0o 180 o 270 o 90 o cos x o = -1 x o = 180 o

Higher Outcome 2 Example 6 Type 5 : Solving the equation 3sin 2 x + 2sin x - 1 = 0 in the range 0 o to 360 o Solving Trig Equations 3x – 1 = 0 x o = 19.5 o and o x o = 90 o Let x = sin x We have 3x 2 + 2x - 1 = 0 (3x – 1)(x + 1) = 0Factorise x = 1/3 x – 1 = 0 x = 1 sin x = 1/3sin x = 1 C A S T 0o0o 180 o 270 o 90 o C A S T 0o0o 180 o 270 o 90 o

Higher Outcome 1 An identity is a statement which is true for all values. eg 3x(x + 4) = 3x x eg(a + b)(a – b) = a 2 – b 2 Trig Identities (1)sin 2 θ + cos 2 θ = 1 (2)sin θ = tan θ cos θ θ ≠ an odd multiple of π/2 or 90°. Trig Identities

Higher Outcome 1 Reason θoθo c b a a 2 +b 2 = c 2 sin θ o = a/c cos θ o = b/c (1) sin 2 θ o + cos 2 θ o = Trig Identities

Higher Outcome 1 = tan  sin 2 θ + cos 2 θ = 1 sin 2 θ = 1 - cos 2 θ cos 2 θ = 1 - sin 2 θ Simply rearranging we get two other forms Trig Identities

Higher Outcome 1 Example1 sin θ = 5 / 13 where 0 < θ < π / 2 Find the exact values of cos θ and tan θ. cos 2 θ = 1 - sin 2 θ = 1 – ( 5 / 13 ) 2 = 1 – 25 / 169 = 144 / 169 cos θ = √ ( 144 / 169 ) = 12 / 13 or - 12 / 13 Since θ is between 0 < θ < π / 2 then cos θ > 0 Socos θ = 12 / 13 tan θ = sinθ cos θ = 5 / 13 ÷ 12 / 13 = 5 / 13 X 13 / 12 tan θ = 5 / 12 Trig Identities

Higher Outcome 1 Given that cos θ = -2 / √ 5 where π< θ < 3 π / 2 Find sin θ and tan θ. sin 2 θ = 1 - cos 2 θ = 1 – ( -2 / √ 5 ) 2 = 1 – 4 / 5 = 1 / 5 sin θ = √( 1 / 5 ) = 1 / √ 5 or - 1 / √ 5 Since θ is between π< θ < 3 π / 2 sinθ < 0 Hence sinθ = - 1 / √5 tan θ = sinθ cos θ = - 1 / √ 5 ÷ -2 / √ 5 = - 1 / √ 5 X - √5 / 2 Hence tan θ = 1 / 2 Trig Identities

Graphs & Functions Strategies Higher Maths Click to start

Graphs & Functions Higher Graphs & Functons The following questions are on Non-calculator questions will be indicated Click to continue You will need a pencil, paper, ruler and rubber.

Hint Quit Previous Next The diagram shows the graph of a function f. f has a minimum turning point at (0, -3) and a point of inflexion at (-4, 2). a) sketch the graph of y = f (- x ). b) On the same diagram, sketch the graph of y = 2 f (- x ) Graphs & Functions Higher a) Reflect across the y axis b) Now scale by 2 in the y direction

Hint Quit Previous Next Graphs & Functions Higher The diagram shows a sketch of part of the graph of a trigonometric function whose equation is of the form Determine the values of a, b and c a is the amplitude: a = 4 b is the number of waves in 2  b = 2 c is where the wave is centred vertically c = 1 2a 1 in  2 in 2  1

Hint Quit Previous Next Graphs & Functions Higher Functions and are defined on suitable domains. a)Find an expression for h ( x ) where h ( x ) = f ( g ( x )). b)Write down any restrictions on the domain of h. a) b)

Hint Quit Previous Next Graphs & Functions Higher a) Express in the form b) On the same diagram sketch i)the graph of ii)the graph of c) Find the range of values of x for which is positive a) b) c) Solve: 10 - f(x) is positive for -1 < x < 5

Hint Quit Previous Next Graphs & Functions Higher The graph of a function f intersects the x -axis at (– a, 0) and ( e, 0) as shown. There is a point of inflexion at (0, b ) and a maximum turning point at ( c, d ). Sketch the graph of the derived function m is + m is - f(x)

Hint Quit Previous Next Graphs & Functions Higher Functions f and g are defined on suitable domains by and a)Find expressions for: i) ii) b)Solve a) b)

Hint Quit Previous Next Graphs & Functions Higher The diagram shows the graphs of two quadratic functions Both graphs have a minimum turning point at (3, 2). Sketch the graph of and on the same diagram sketch the graph of y=g(x) y=f(x)

Hint Quit Previous Next Graphs & Functions Higher Functions are defined on a suitable set of real numbers. a) Find expressions for b) i)Show that ii) Find a similar expression for and hence solve the equation a) b) Now use exact values Repeat for ii) equation reduces to

Hint Quit Previous Next Graphs & Functions Higher A sketch of the graph of y = f(x) where is shown. The graph has a maximum at A and a minimum at B(3, 0) a) Find the co-ordinates of the turning point at A. b) Hence, sketch the graph of Indicate the co-ordinates of the turning points. There is no need to calculate the co-ordinates of the points of intersection with the axes. c) Write down the range of values of k for which g(x) = k has 3 real roots. a) Differentiate for SP, f(x) = 0 when x = 1t.p. at A is: b) Graph is moved 2 units to the left, and 4 units up t.p.’s are: c) For 3 real roots, line y = k has to cut graph at 3 points from the graph, k  4

Hint Quit Previous Next Graphs & Functions Higher a) Find b) If find in its simplest form. a) b)

Hint Quit Previous Next Graphs & Functions Higher Part of the graph of is shown in the diagram. On separate diagrams sketch the graph of a)b) Indicate on each graph the images of O, A, B, C, and D. a) b) graph moves to the left 1 unit graph is reflected in the x axis graph is then scaled 2 units in the y direction

Hint Quit Previous Next Graphs & Functions Higher Functions f and g are defined on the set of real numbers by a) Find formulae for i) ii) b) The function h is defined by Show that and sketch the graph of h. c) Find the area enclosed between this graph and the x-axis. a) b) c) Graph cuts x axis at 0 and 1Now evaluate Area

Hint Quit Previous Next Graphs & Functions Higher The functions f and g are defined on a suitable domain by a) Find an expression for b) Factorise a) Difference of 2 squares Simplify b)

You have completed all 13 questions in this section Back to start Quit Previous Graphs & Functions Higher

30°45°60° sin cos tan1 Table of exact values Previous Graphs & Functions Higher